a) ([tex]\frac{-1}{2} + \frac{\sqrt{3} }{2}i[/tex])³ . ([tex]\sqrt{3} + i[/tex])²

→ Resolveremos a Primeira Parte:

([tex]\frac{-1}{2} + \frac{\sqrt{3} }{2}i[/tex]) . ([tex]\frac{-1}{2} + \frac{\sqrt{3} }{2}i[/tex]) . ([tex]\frac{-1}{2} + \frac{\sqrt{3} }{2}i[/tex]) = [tex]\frac{1+\sqrt{3}i }{4}[/tex]

→ Resolveremos a Segunda Parte:

[tex](\sqrt{3}+i) ^{2}[/tex] = [tex]\sqrt{9}+\sqrt{3i} + \sqrt{3i} + i^{2}\\[/tex] => [tex]3 + 2 \sqrt{3i} -1[/tex] => [tex]2+ 2\sqrt{3i}[/tex]

→ Agora Multiplicaremos com a segunda parte:

=> [tex](\frac{1+\sqrt{3}i }{4}) . (2+ 2\sqrt{3i} )[/tex]

=> [tex]2+2\sqrt{3i } + 2\sqrt{3i } +2\sqrt{9} i^{2}[/tex]

=> [tex]\frac{2+4\sqrt{3} i-2.3}{4}[/tex]

=> [tex]\frac{-4+4\sqrt{3} i}{4}[/tex]  (Simplifica)

====>   [tex]z^{3} . w^{2}[/tex] =  [tex]-1+\sqrt{3} i[/tex]

b) [tex]\frac{\frac{-1}{2} + \frac{\sqrt{3} }{2} i}{\sqrt{3}+1 }[/tex] , Primeiro temos que multiplicar pelo conjugado, isto é,

=> [tex]\frac{\frac{-1}{2} + \frac{\sqrt{3} }{2} i .(\sqrt{3}-i) }{\sqrt{3}+1. (\sqrt{3}-1)}[/tex]

=>  [tex]\frac{\frac{-1+\sqrt{3}i .(\sqrt{3} -i}{2} }\sqrt{9} -\sqrt{3}i+\sqrt{3}i-i^{2}[/tex]

=>  [tex]\frac{-\sqrt{3}+i+\sqrt{9}-\sqrt{3} i^{2} }{4}[/tex]

=>  [tex]\frac{-\sqrt{3}+i+3+\sqrt{3} }{4}[/tex]  (Corta as raízes)

=====> [tex]\frac{3+i}{4}[/tex]